Compare two arrays using Array_diff () why only a different element is returned?
I take out the values of the same fields in the different tables in the database in two different arrays, using Array_diff () to find the elements in array 1 instead of the array 2, why only a different element is returned?
The code is as follows:
Error_reporting (E_all&~e_notice);
Header (' content-type:text/html; Charset=utf-8 ');
Date_default_timezone_set (' Asia/shanghai ');
Require_once ("init.php");
$sql = "SELECT * from Dev_property";
$res =mysql_query ($sql) or Die (Mysql_error ());
$num _rows=mysql_num_rows ($res);
while ($sn =mysql_fetch_array ($res))
{
$prime =array ();
$prime []= $sn ["Serial_number"];
for ($i =0; $i <>
{
echo $prime [$i];
echo "
";
}
}
echo "
";
$starttime = ' 08:25:00 ';
$endtime = ' 23:35:00 ';
$sql = "SELECT * from Data where time>= '". Date (' y-m-d '). ' '. $starttime. ' and time<= '. Date (' y-m-d '). ' '. $endtime. ' ";
$result =mysql_query ($sql) or Die (Mysql_error ());
while ($date =mysql_fetch_array ($result))
{
$num _rows1=mysql_num_rows ($result);
$a =array ();
$a []= $date ["Serial_number"];
for ($i =0; $i <>
{
echo $a [$i];
echo "
";
}
echo "
";
}
$diff =array_diff ($prime, $a);
for ($i =0; $i <>
echo $diff [$i];
echo "
";
echo count ($diff);
echo "
";
?>
Execution Result:
Array $prime
862118028862461
862118028862462
862118028862463
862118028862464
Array $ A
862118028862461
Results
862118028862464
1
What I meant to return was,
862118028862462
862118028862463
862118028862464
Now only returns 862118028862464, why?
------to solve the idea----------------------
You are $diff =array_diff ($prime, $a); Print the array before
Print_r ($prime);
Print_r ($a);
$diff =array_diff ($prime, $a);
Look what it is.
